The denominator of f(x) is (x²-5x). When we factorize this, we get,
(x²-5x)=x(x-5)
Since denominator becomes 0 at x=0 and x=5, f(x) is discontinuous at x=0 and x=5.
The denominator of g(x) is (x²-6x). When we factorize this, we get,
(x²-6x+8)=x(x-6)
Since denominator becomes 0 at x=0 and x=6, f(x) is discontinuous at x=0 and x=6.

z=x+iy
In general limits are discussed at origin, if nothing is specified.
f(x, y)=(x+iy)²+2(x+iy)
f(x, y)=x²-y²+2xiy+2x+2iy
Left limit: \(\lim_{x \to 0 \\ y \to 0}\)x²-y²+2xiy+2x+2iy
\(\lim_{y \to 0}\)-y²+2iy
=0
Right limit: \(\lim_{y \to 0 \\ x \to 0}\)x²-y²+2xiy+2x+2iy
\(\lim_{x \to 0}\)x²+2x
=0
Both limits are equal, refore function is continuous. To check differentiability,
f?(z)=\(\lim_{\delta z \to 0}\frac{f(z+\delta z)-f(z)}{\delta z}\) should exist.
\(\lim_{\delta z \to 0}\frac{(z+\delta z)^2+2(z+\delta z)-(z^2+2z)}{\delta z}\)
\(\lim_{\delta z \to 0}\frac{z^2+(\delta z)^2+2z(\delta z)+2z+2(\delta z)-z^2-2z}{\delta z}\)
=2z+2
Since f?(z) exists, function is differentiable as well.

Writing z=x+iy, we get L as 2x?3y?1 and S as x²+y²?6, a point z₁ lies in smaller region if L₁>0 and S₁<0. ∴ (2+3i/4) and (1/4-i/4) lie in smaller region.

z⁵-5z+a=0 ⇒ z⁵-5z=-a ⇒ z(z-5^1/4)(z+5^1/4)(z²+5^1/2)=-a f'(z)=5z⁴?5=0 ⇒ (z²+1)(z²-1)=0 ⇒ (z-1)(z+1)(z²+1)=0 ⇒ α has exactly one possible real value for a>4 and exactly three possible real values for -4

Case 1: \(\lim_{x \to 0 \\ y \to 0}\frac{3xy+3y+3x}{5x+5y}\)
\(\lim_{y \to 0}\frac{3y}{5y}\)=\(\frac{3}{5}\)
Case 2: \(\lim_{y \to 0 \\ x \to 0}\frac{3xy+3y+3x}{5x+5y}\)
\(\lim_{x \to 0}\frac{3x}{5x}\)=\(\frac{3}{5}\)
Case 3: Along a linear path, \(\lim_{y \to mx \\ x \to 0}\frac{3xy+3y+3x}{5x+5y}\)
\(\lim_{x \to 0}\frac{3mx^2+3mx+3x}{5x}\)=\(\frac{3(m+1)}{5}\)
Since, cases are not equal, f(x, y) is discontinuous at x=0.

\(\frac{x^2-81}{x-9}\)=\(\frac{(x-9)(x+9)}{x-9}\)=(x-9) which is continuous at x = 9.
\(\frac{x^3-20x^2+117x-162}{x-9}\)=\(\frac{(x-9)^2(x-2)}{(x-9)}\)=(x-9)(x-2) which is continuous at x = 9.
\(\frac{x^2-11x+18}{x(x-9)}\)=\(\frac{(x-2)(x-9)}{x(x-9)}\)=\(\frac{(x-2)}{x}\) which is continuous at x = 9.
\(\frac{x^2-13x-90}{x-9}\)=\(\frac{(x-18)(x+5)}{x-9}\) which is discontinuous at x = 9, since denominator becomes 0.

f(z)=az²+b, with a, b of same sign ⇒ f(f(z))=a(az²+b)²+b
If z∈R or iz∈R ⇒ z²∈R ⇒ f(z)∈R ⇒ f(f(z))≠0 ⇒ Hence real or purely imaginary number cannot satisfy f(f(z))=0.

Let y=1/(1-z) ⇒ z=1-1/y
|z|=1 ⇒ |1-1/y|=1 ⇒ |y-1|=|y| ⇒ locus of y is perpendicular bisector of line segment joining 0 and 1 ⇒ arg y ∈(-π/2, π/2).

When we substitute x=0, x=5 or x=-2 in f(x), we get denominator as 0. This is because se are points of discontinuity.
At x=2, we get f(2)=\(\frac{-1}{24}\) which is continuous.

z=x+iy
In general limits are discussed at origin, if nothing is specified.
f(x,y)=(x+iy)²
Left limit: \(\lim_{x \to 0 \\ y \to 0}\)(x+iy)²
\(\lim_{y \to 0}\)i²y²
=0
Right limit: \(\lim_{y \to 0 \\ x \to 0}\)(x+iy)²
\(\lim_{x \to 0}\)x²
=0
Both limits are equal, refore function is continuous. To check differentiability,
f?(z)=\(\lim_{\delta z \to 0}\frac{f(z+\delta z)-f(z)}{\delta z}\)should exist.
\(\lim_{\delta z \to 0}\frac{(z+\delta z)^2-(z)^2}{\delta z}\)
\(\lim_{\delta z \to 0}\frac{2z(\delta z)+(\delta z)^2}{\delta z}\)
=\(\lim_{\delta z \to 0}\)(2z+(δz))
=2z
Since f?(z) exists, function is differentiable as well.